# Intuitively understanding Riemann surfaces

I'm looking at the Riemann surface of $f(z) = z^{1/2}$ so the set $\{(z,w) \in \mathbb{C}^2 : w^2 = z \}$. I understand that the point of the riemann surface is to understand this multi-valued function.

Now I visualise this as taking two copies of the complex plane both with a slit in them (negative real axis say) and then I put them on top of each other, flip the top one and then sort of join them up along the slit and that's fine and I can see that that will form some surface where the local coordinates are given by projection onto the z or w axis. But here I am just dealing with the domain of the function $f(z)$.

My difficulty is when I think of the analogous situation with say $\{(x,y) \in \mathbb{R}^2 : y^2 = x\}$ this "surface" is just a line and I can see how given a point on this "line" we have an associated pair $(x,y)$. Now when I look at the Riemann surface described above all I see is the domain of the function $f(z)$ I can't see how each point of the surface is in anyway related to a pair $(z,w)$.

I realise this is quite a vague question and I am struggling to put my frustration with this concept into words - I hope this makes sense! I've now added a (terrible) diagram which may help to illustrate my issue: Thanks

## 1 Answer

To "see" $z$ and $w$ you need an embedding, not just the abstract description by gluing slit planes. :)

Below is the Riemann surface of $w = \sqrt{z}$, orthogonally projected into the $3$-dimensional space containing the $z$ (complex) line and the real axis of $w$. The blue axes span the $z$ line. The two points above/below a $z$-value have as their vertical coordinate the real parts of the two square roots of $z$. The shaded branch and the transparent branch are each defined on the complement of the non-negative reals. Each "interpolates" between the positive (yellow) and negative (red) real square root. (It's not easy to "see" the $w$-values here, since the imaginary part of $w$ has been projected out.) • Thank you for this great answer - this is really clear to me now. I can see my confusion was that I was thinking of the two surfaces I drew as already being embedded in $\mathbb{R}^3$ whereas of course this is not the case! I just drew the abstract topological space. – Wooster Oct 29 '14 at 13:12
• @Wooster: You're welcome; glad the picture was helpful. :) – Andrew D. Hwang Oct 29 '14 at 13:50